Pell’s Equation

The Puzzle

Here is a puzzle that I will use to illustrate the use and solution of Pell’s equation.

I am studying a six digit item number in a very long list of consecutively numbered items.

To my surprise I have found that if I add this item number to a certain number of the immediately lower item numbers, the result is the same as the sum of the same number of immediately higher item numbers.

Also, if I add the square of this item number to the squares of a different number of immediately lower item numbers, the result is the same as the sum of the squares of the same number of immediately higher item numbers.

What is the number that I am studying?

Analysis

Let \(i\) be the item number being considered and \(m\) be the number of immediately lower and higher item numbers being added. So we have: \[(i-m) +\cdots +(i-1) + i = (i + 1) +\cdots + (i+m)\]Simplifying this gives \[i=m(m+1)\]Now repeating this for \(n\) squared values: \[(i-n)^2 +\cdots +(i-1)^2 + i^2 = (i + 1)^2 +\cdots + (i+n)^2\] which simplifies to:\[i=2n(n+1)\]Now eliminating the item number \(i\) we obtain: \[m(m+1)= 2n(n+1)\] which can be recast as: \[(2m+1)^2 – 2(2n+1)^2 = -1\]

Quadratic Diophantine Equations

To solve this we now need a diversion to study solutions for equations of the form: \[x^2-Dy^2=N\] for \(D\) a non-square positive integer, \(c\) an integer and both \(x\) and \(y\) integer. This is known as a quadratic diophantine equation but is also known as Pell’s equation when \(N=1\) (it was mistakenly attributed it to Pell by Euler). Before considering a full solution, it is useful to set out how we can develop a whole sequence of solutions once we have a single solution. If we have a solution \((x_k, y_k)\) such that \[x_k^2 – Dy_k^2=N\]and \((r, s)\) values such that:\[r^2-Ds^2=1\]then the identity:\[(x_k^2 – Dy_k^2)(r^2-Ds^2) = (rx_k\pm Dsy_k) ^2- D(sx_k\pm ry_k)^2\]shows that\[(x_{k+1},y_{k+1}) = (rx_k\pm Dsy_k, sx_k\pm ry_k)\]is also a solution.

For our puzzle with \(D=2\) an initial solution for \((x,y)\) is \((1)^2 – 2(1)^2=-1\) and one for \((r,s)\) is \((3)^2 – 2(2)^2=1\). So we can now develop a sequence of solutions as:\[(x_{k+1}, y_{k+1}) = (3x_k+4y_k,2x_k+3y_k)\]These solutions are provided by this Python program